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\begin{align*} \left|\left(1+\frac{a+b}{n}\right)^{n}-\left(1+\frac{a}{n}\right)^{n}\left(1+\frac{b}{n}\right)^{n}\right| & =\left|\left(1+\frac{a+b}{n}\right)^{n}-\left[\left(1+\frac{a}{n}\right)\left(1+\frac{b}{n}\right)\right]^{n}\right|\\ & =\left|\left[\left(1+\frac{a+b}{n}\right)-\left(1+\frac{a}{n}\right)\left(1+\frac{b}{n}\right)\right]\sum_{k=0}^{n-1}\left(1+\frac{a+b}{n}\right)^{k}\left(1+\frac{a}{n}\right)^{n-1-k}\left(1+\frac{b}{n}\right)^{n-1-k}\right|\\ & =\left|-\frac{ab}{n^{2}}\sum_{k=0}^{n-1}\left(1+\frac{a+b}{n}\right)^{k}\left(1+\frac{a}{n}\right)^{n-1-k}\left(1+\frac{b}{n}\right)^{n-1-k}\right|\\ & \le\frac{\left|ab\right|}{n^{2}}\sum_{k=0}^{n-1}\left(1+\frac{\left|a+b\right|}{n}\right)^{k}\left(1+\frac{\left|a\right|}{n}\right)^{n-1-k}\left(1+\frac{\left|b\right|}{n}\right)^{n-1-k}\\ & \le\frac{\left|ab\right|}{n^{2}}\sum_{k=0}^{n-1}\left(1+\frac{\left|a+b\right|}{n}\right)^{n}\left(1+\frac{\left|a\right|}{n}\right)^{n}\left(1+\frac{\left|b\right|}{n}\right)^{n}\\ & =\frac{\left|ab\right|}{n}\left(1+\frac{\left|a+b\right|}{n}\right)^{n}\left(1+\frac{\left|a\right|}{n}\right)^{n}\left(1+\frac{\left|b\right|}{n}\right)^{n} \end{align*}
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Thu, 06 Feb 2025 08:14 GMT